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I'm looking for some streamlined ways to verify that certain maps are topological embeddings, since these pop up all the time in examples. For example, I was able to prove that the helix map $\mathbb{R} \to \mathbb{R}^3$ given by $t \mapsto (\cos2\pi t, \sin 2\pi t, t)$ is an embedding by checking that it's closed (I proved this by showing that the complements of open intervals are taken to closed sets in $\mathbb{R}^3$) but this map is so obviously an embedding in my intuition that I feel like there must be a cleaner way to prove it. I'm looking for something similar to the result that any injective continuous map of a compact space into a Hausdorff space is automatically an embedding.

In general, what are some easy techniques to prove that a map is an embedding in practice? I always find myself trying to prove that the map is open or closed, which isn't always that straightforward.

Nick A.
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  • The simplest examples to check are going to be the ones where the tools of calculus can be brought to bear. The following may be useful to you: https://math.stackexchange.com/questions/1116124/embeddings-are-precisely-proper-injective-immersions – Mike F Aug 10 '20 at 16:57
  • @MikeF One answer in the post you linked mentioned without proof that closed embeddings are the same as embeddings with closed image. Why is this true? – Nick A. Aug 10 '20 at 17:02
  • Let $f : X \to Y$ be an embedding. Suppose $f$ is closed as a map from $X$ to $Y$. Then, because $X$ is closed in itself, we have that $f(X)$ is closed in $Y$. Conversely, suppose that $f(X)$ is closed in $Y$. Let $A$ be a closed subset of $X$. Firstly, $f(A)$ is closed in $f(X)$, since $f$ is a homeomorphism of $X$ onto $f(X)$. Since closed subsets of closed subsets are closed, we also have $f(A)$ closed in $Y$. – Mike F Aug 10 '20 at 17:12
  • Though I should add that, under that link, "embedding" can be assumed to mean "smooth embedding", which is a more restrictive type of mapping than a "continuous embedding" – Mike F Aug 10 '20 at 17:16

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The given function is continuous.
As it is a bijection from R onto a helix it has an inverse.
Show the inverse is continuous.
From that, it's apparent it is an embedding.

William Elliot
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    Right, the inverse is just the restriction of the projection $(x,y,z) \mapsto z$ to the helix, which is clearly continuous - that's a much easier way to prove it. – Nick A. Aug 10 '20 at 16:59